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nadia.rasul
Course Students
 
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Book #5, pg.125 question 7

by nadia.rasul Fri Oct 05, 2012 11:54 pm

I don't understand the solution to this question. I was trying to do the math for it and got confused. Wouldn't the probability of getting more heads than tails be 3/5 (1 tails + 4 heads, 2 tails + 3 heads or all 5 heads) which is not the same as quantity B?
tommywallach
Manhattan Prep Staff
 
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Re: Book #5, pg.125 question 7

by tommywallach Sun Oct 07, 2012 11:10 pm

Hey Nadia,

I don't have the books in front of me, but it sounds like you're asking for the probability that one will get more heads than tails when flipping a coin 5 times.

The easiest way to solve is to imagine that the odds of getting more heads than tails must be exactly the same as getting more tails than heads. There can't be a difference, given that there's never a better chance of getting heads than tails.

Or did I misunderstand your question?

-t

P.S. You seem to be counting 1T/4H as one possibility, where as 2T/3H is another possibility. But those two things don't have an equal probability of happening:

The ways to get 1T/4H: THHHH, HTHHH, HHTHH, HHHTH, HHHHT
The ways to get 2T/3H: TTHHH, THTHH, THHTH, THHHT, HTTHH,
HTHTH, HTHHT, HHTTH, HHTHT, HHHTT

See, there are only 5 ways to get 1T/4H, but 10 ways to get 2T/3H.

-t